Question

find the value of k for which the quadratic equation (3k+1)x2 + 2 (k+1) x+1=0 has equal roots also find the roots.
please solve this question...............
don't copy...........​

Answer 1

Answer:

${ \large \bold \orange{ \underline{answer \: \: = value \: of \: k \: is \: 0 \: and \: 1}}}$

Step-by-step explanation:

$\mathfrak{ \large \bold \red{ \underline{ \underline{question}}}} \\ { \large{(3k + 1) {x}^{2} + 2(k + 1)x + 1 = 0 }} \\ \\ { \large \green{ \underline{to \: find}}} \\ { \large{value \: of \: k \: and \: roots}} \\ \\ { \large \bold \blue{ \underline{solution :}}} \\ { \large{a = (3k + 1) \: and \: b = 2(k + 1) \: and \: c = 1}} \\ { \large{ \underline{find \: dicriminant \: d = {b}^{2} - 4ac }}} \\ { \large{(2(k + 1)) {}^{2} - 4(3k + 1) \times 1 }} \\ { \large{4 {k}^{2} + 4 + 8k - 12k - 4 }} \\ { \large{4k(k - 1)}} \\ \\ { \large \pink{ \underline{ : since \: roots \: are \: real \: and \: equal}}} \\ { \large{ \underline{put \: d = 0}}} \\ \\ { \large{4k(k - 1) = 0}} \\ { \large{k = 0 \: and \: k - 1 = 0}} \\ { \large{k = 0 \: and \: k = 1}} \\ \\ { \large \bold \orange{ \underline{answer \: \: = value \: of \: k \: is \: 0 \: and \: 1}}}$